Use the four-step procedure for solving variation problems given on page 447 to solve Exercises 1–10. y varies jointly as a and b and inversely as the square root of c. y = 12 when a = 3, b = 2, and c = 25. Find y when a = 5, b = 3 and c = 9.

Verified step by step guidance

1. Step 1: Identify the type of variation described. Since y varies jointly as a and b and inversely as the square root of c, we can write the variation equation as: \[y = k \cdot a \cdot b \div \sqrt{c}\] where \(k\) is the constant of variation.
2. Step 2: Use the given values \(y = 12\), \(a = 3\), \(b = 2\), and \(c = 25\) to find the constant \(k\). Substitute these values into the equation: \[12 = k \cdot 3 \cdot 2 \div \sqrt{25}\]
3. Step 3: Simplify the right side of the equation to solve for \(k\). Calculate the square root of 25 and multiply the known values: \[12 = k \cdot 6 \div 5\] Then solve for \(k\) by isolating it on one side.
4. Step 4: Once \(k\) is found, write the complete variation equation with \(k\) included: \[y = k \cdot a \cdot b \div \sqrt{c}\]
5. Step 5: Use the new values \(a = 5\), \(b = 3\), and \(c = 9\) to find the new value of \(y\). Substitute these values and the constant \(k\) into the equation and simplify: \[y = k \cdot 5 \cdot 3 \div \sqrt{9}\]

Key Concepts

Joint and Inverse Variation

Joint variation occurs when a variable depends on the product of two or more variables, while inverse variation means a variable depends on the reciprocal of another. In this problem, y varies jointly as a and b, and inversely as the square root of c, combining both types of variation in one relationship.

Formulating the Variation Equation

To solve variation problems, write an equation expressing the dependent variable in terms of the independent variables and a constant of proportionality. Here, y = k * a * b / √c, where k is found using given values, enabling calculation of y for new inputs.

Solving for the Constant of Proportionality

Use the provided values of y, a, b, and c to substitute into the variation equation and solve for the constant k. This step is crucial because k allows the equation to model the specific situation accurately before finding y for new variable values.